17 MEROMORPHIC FUNCTIONS 247

J(a;y)g(a)a>(a;fl

/y J \^J

a finite number of terms only being ^0 on the right-hand side.

This follows at once from the theorem of residues, for the residue of
gflf at a point a e S u T is the product of g(a) by the residue off'ff at the
point a.

(9.17.2) "With the assumptions of (9.17.1) let t-*y(t) (f el) be a circuit in
A - (S u T). IfT is the circuit t ->f(y(t))9 then

aeS u T

For it follows at once from (8.7.4) that

L^'LjiH)**'

hence the result is a particular case of (9.17.1) for g = 1.

(9.17.3) (Rouche's theorem) Let A c C be a simply connected domain,
/, # rv^o analytic complex valued functions in A. LeŁ T be the (at most de-
numerable) set of zeros off, T' the set of zeros off+g in A, y a circuit in
A - T, defined in an interval I. Then, if \g(z)\ < \f(z)\ in y(I), the function
f+g has no zeros on y(I), and

Zj(a;y)a>(a;f)= % j(b; y)co(b;f+ g).

aeT beT

The first point is obvious, since /(z) + g(z) = 0 implies |/(z)| = \g(z)\. The
function h = (f-\- g)/fis defined in A — T and meromorphic in A; we have

Using (9.17.2), all we have to prove is that the index of 0 with respect to
the circuit F : t -» h(y(t)) is 0. As g/f is continuous and finite in the
compact set y(I), it follows from (3.17.10) and the assumption that
r = sup \g(z)lf(z)\ < 1. In other words, F is in the ball \z — 1| < r, and as

zey(I)

0 is exterior to the ball, the result follows from (9.8.5). A and co(d) =h, then f(z) = (z - dffa(z) in a set 0 < \z - a\ < r,